Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M CH3COOH(aq) with 0.150 M KOH(aq). The ionization constant for HClO is Ka = 1.8 x 10-5. Just give the number to 2 decimal places.

A) Before addition of any KOH: pH =

B) After addition of 25 mL of KOH: pH =

C) After addition of 50 mL of KOH: pH =

D) After addition of 60 mL of KOH: pH

Answer 1

This is a titration, so we will have different pH at different Volumes:

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

formulas we will need:

M = moles / volume

pH = -log[H+]

pKa = -log(Ka) = -log(1.8*10^-5) = 4.75

Let

CH3COOH be "HA" for simplicity so it will dissociate as follows:

CH3COOH <--> H+ + CH3COO-

HA <--> H+ + A-

Ka = [H+][CH3COO-]/[CH3COOH]

Ka = [H+][A-]/[HA]

now...

a) No base, only weak acid

initially HA -> H+ + A-

Ka = [H+][A-]/[HA]

Assume [H+] = [A-] = x; due to stoichiometry

[HA] = M – x (x accounts for the dissociated acid in equilibrium)

Substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

1.8*10^-5 = x*x/(0.150-x)

This is quadratic equation

x =0.00163

For pH

pH = -log(H+)

pH =-log(0.00163 )

pH in a = 2.08990

b) when Vbas = 25 ml

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

pH = pKa + log (A-/HA)

where;

A- is the conjugate and HA the potonated acid:

initially

mmol of acid = MacidVacid = 50*0.150 = 7.5 mmol of acid

mmol of base = MbaseVbase = 25*0.150 = 3.75 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 7.5 - 3.75 = 3.75 mmol

mmol of conjguate left = 0 + 3.75 = 5

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.75 + log (3.75 /3.75 )

pH = 4.75

c) for 50 ml

Addition of Same quantitie of Acid/Base

This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium

We will need Kb

Ka*Kb = Kw

And Kw = 10^-14 always at 25°C for water so:

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

Now, proceed to calculate the equilibrium

H2O + A- <-> OH- + HA

Then K equilibrium is given as:

Kb = [HA][OH-]/[A-]

Assume [HA] = [OH-] = x

[A-] = M – x

M = mmol of conjugate / Total V = (7.5) / (50+50) = 0.075 M

Substitute in Kb

5.55*10^-10 = [x^2]/[0.075-x]

x = 6.45*10^-6

[OH-] =6.45*10^-6

Get pOH

pOH = -log(OH-)

pOH = -log (6.45*10^-6) = 5.190

pH = 14-pOH = 14-5.190= 8.81

e) Addition of base

There will be finally an Excess of Base!

mol of acid < mol of base

Calculate pOH directly

[OH-] = M*V / Vt

mmol of acid = MV = 50*0.150= 7.5

mmol of base = MV = 60*0.150 = 9

therefre,

mmol of strng base left = 9-7.5 = 1.5 mmol

Vtotal = 60+50= 110mL

[OH-] = 1.5 /110 = 0.01363636

pOH = -log(OH-)

pOH = -log(0.01363636) = 1.86530

pH = 14-pOH = 14-1.86530= 12.134

pH = 12.134